Question

The normal to the curve $x^2=4y$ passing $(1,2)$ is

• A. $x+y=3$
• B. $x-y=3$
• C. $x+y=1$
• D. $x-y=1$
• E. answer required

Answer 1

Answer: (A)

$x+y=3$

Given, $x^2=4y$

Slope of tangent to the curve will be $4\frac{dy}{dx}=2x$
$\Rightarrow \frac{dy}{dx}=\frac{x}{2}$
So slope of normal will be $-\frac{dx}{dy}=\frac{-2}{x}$
Let the point on the curve be $(h,k)$
So slope of normal will be $\frac{-2}{h}$
Equation of normal is given by $(y-k)=\frac{-2}{h}(x-h)$
It is given that the normal passes through $(1,2)$
$\Rightarrow (2-k)=\frac{-2}{h}(1-h)$.............(1)
Also, $(h,k)$ lies on the curve so $h^2=4k$
From (1) we have
$\frac{h^3}{4}=2h+2-2h=2$
$\Rightarrow h^3=8$
$\Rightarrow h=2$
$k=\frac{h^2}{4}\Rightarrow k=1$
So the equation will be
$y-1=\frac{-2}{2}(x-2)$
$x+y=3$